ΔG°=−RTlnKp
Partial pressure of oxygen ;11.6 Pa = 0.0001144831 ATM
First write the equilibrium constant in form of pressure = product / reactnats
Kp = 1/ (O2)^1/2
FOR SOLID PRESSURE IS EQUAL = 1
Kp = 1/ 0.0001144831)^1/2
= 1/0.0107
= 93.46
Here T = 823 K AND R = 8.314 J / (mol. · K)
ΔG°=−RTlnKp
= - 8.314 J / (mol. · K) *823 K ln 93.46
=--31049.5865 J/ mole *1 kj/1000 J
= - 31.0KJ/ MOLE
THUS THE CORRECT ANSWER IS a)
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